По каждой из шести одинаковых тонких непроводящих ...

問題

По каждой из шести одинаковых тонких непроводящих пластин равномерно распределен одинаковый заряд. Используя эти пластины в качестве граней куба, собрали полный куб, после чего стали измерять модуль напряженности электрического поля в центре куба, поочередно удаляя пластины. После удаления первой пластины измерения дали значение Е1=Е, а после удаления всех шести пластин оказалось Е6=0. Восстановите четыре промежуточных результата измерений Е2, Е3, Е4, Е5. какие именно пластины удаляли на каждом этапе-неизвестно. Представьте все возможные варианты ответа

思考

Understanding electric field at cube’s center

To figure out the electric field at the center of a cube with six uniformly charged faces, I consider symmetry. Opposite faces create equal and opposite fields that should cancel out. Each face’s electric field points away from it. At the cube’s center, these fields should cancel, but I’ll explore this further.

Calculating electric field after removing plates

When a plane at x=+a/2 is removed, the field at the origin points away from the plane, causing the field contribution from opposite plates to cancel out. With all six faces, the net field is zero. As plates are removed, we’ll measure the field’s magnitude at the center to determine E2 to E5, depending on the remaining plates. Each face has the same magnitude field, E0.

Relating field components to plate removal

We don’t need the absolute value of E0 for this calculation, just the relative field. I’ll compute the net field by considering the vector sum of contributions from the remaining faces. Each face’s contribution is along the direction of its normal, and each pair of opposite faces will cancel out. The net field is given by E0 times the difference between the number of remaining faces along each axis.

Determining magnitude after each plate removal

Each component of the electric field is a multiple of E0, depending on whether one face of a pair is missing. If both faces are absent or present, the contribution is zero. After removing a plate, the configuration changes and the field's magnitude depends on how many unpaired faces remain along each axis. After removing one plate, the magnitude equals E0. Starting from E = E1, we can continue calculating the field based on which faces remain along each axis.

Understanding the effect of plate removal on the field

When removing a face, it unpairs the corresponding axis unless its opposite face is later also removed, which resets the pairing. The field magnitude depends on how many axes have one face remaining. After each removal, the state of each axis toggles, and I count how many axes have exactly one face removed. This gives the field magnitudes as E, E*sqrt2, and so on.

Enumerating possible field magnitudes after plate removal

I need to track how many faces are removed from each axis (r_x, r_y, r_z), with each axis having two removals. The field magnitude depends on how many axes have one face remaining (m). After each step, I calculate m(k) based on how many axes have exactly one removed face. At each step, the magnitude varies according to the number of unpaired axes.

Analyzing possible sequences of face removals

Exploring the removal sequence, for each step, I track the number of faces removed from each axis. At each point, the value $m(k)$ depends on how many axes have one face removed. For example, for k=1, only one axis has one removal. By the time all faces are removed, I analyze the possible patterns for $m(k)$ values at k=2, k=3, etc.

Exploring different symbol sequences

I’m considering how the sequence of removals impacts the stepwise magnitudes. At each step, I look at the counts for each axis, identifying cases where face counts change in specific ways. Symmetry between axes helps streamline this, but the sequence still influences potential outcomes for each step.

Breaking down possible sequences and magnitudes

I’m analyzing different possible counts at each step and how they affect the sequence's evolution. Based on the sequence, I can identify key patterns and work out the corresponding magnitudes (E values) for each step. This helps me identify distinct sequences of values and their corresponding magnitudes like E2 through E5.

Presenting all possible variants and logical reasoning

There are five possible variants to consider:

(0, E, 0, E)
(0, E, E√2, E)
(E√2, E, 0, E)
(E√2, E, E√2, E)
(E√2, E√3, E√2, E)

Each variant’s logic works by analyzing magnetism and plate removals. The E2 = 0 scenario is possible when opposite axis pairs are intact, and careful checks make sure the calculation holds. When translating the explanation into Russian, we can include cube face removal logic and reasoning.

Thought for 2m 43s

回答

Рассмотрим куб с центром в начале координат. Поле в центре от каждой равномерно заряженной грани направлено перпендикулярно этой грани (вдоль одной из осей) и имеет одинаковый модуль $E_0$ .

Для каждой пары противоположных граней (например, $x=+a/2$ и $x=-a/2$ ):

если обе грани есть (или обе удалены), их вклады по этой оси взаимно компенсируются $\Rightarrow$ компонента по оси = 0;
если осталась ровно одна грань из пары, появляется компонента по этой оси величины $E_0$ (знак неважен, т.к. меряем модуль).

Значит, модуль поля в центре равен

E_k = E_0\sqrt{m_k},

где $m_k$ — число осей $x,y,z$ , для которых “пара граней нарушена” (удалена ровно одна грань из пары).

После удаления первой пластины $m_1=1$ , и дано $E_1=E$ . Значит $E_0=E$ .

Дальше возможны только $m_k\in\{0,1,2,3\}$ , то есть значения $0,\;E,\;E\sqrt2,\;E\sqrt3$ .
Перебор всех возможных порядков удаления (по сути — последовательностей из трёх пар, каждая встречается дважды) даёт ровно 5 разных вариантов для $(E_2,E_3,E_4,E_5)$ :

Все возможные ответы

(E_2,E_3,E_4,E_5)=(0,;E,;0,;E)

2) $$ (E_2,E_3,E_4,E_5)=(0,\;E,\;E\sqrt2,\;E)

(E_2,E_3,E_4,E_5)=(E\sqrt2,;E,;0,;E)

4) $$ (E_2,E_3,E_4,E_5)=(E\sqrt2,\;E,\;E\sqrt2,\;E)

(E_2,E_3,E_4,E_5)=(E\sqrt2,;E\sqrt3,;E\sqrt2,;E)

(И всегда $E_1=E$, $E_6=0$.) Если хотите, могу для каждого из 5 вариантов привести не один, а полный класс порядков удаления (в терминах “с какой оси снимали грань” / “какие противоположные пары”).

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